let $s = \frac x{y} + \frac y{x}$. The given equation then becomes $s^2+s-2=0 \rightarrow (s+2)(s-1)=0$. If we let $r = \frac x{y}$, then $ s = r + \frac 1{r} = 1$ or $s = r + \frac 1{r} = -2$. The two quadratics then become $r^2-r+1=0$ and $r^2+2r+1=0$. The second one yields a solution of $\frac{x}y = -1$, which violates the inequality since $\frac{x}{y} = -1 \rightarrow x = -y \rightarrow x+y = 0 \not < 0 $. Now we are left with the quadratic of $r^2-r+1=0$, which has no real solutions. Thus there are no solutions.